warm up - city university of new york
TRANSCRIPT
Warm up
Recall, to graph a conic function, you want it in the formparabola: (x− x0)2 = 4p(y − y0) or (y − y0)2 = 4p(x− x0),
ellipse:(x−x0a
)2+(y−y0
b
)2= 1,
hyperbola:(x−x0a
)2 − (y−y0b
)2= 1 or
(y−y0a
)2 − (x−x0b
)2= 1
Without doing any algebraic manipulation, decide whether each ofthe following conic sections is a parabola, ellipse, or hyperbola.Can you tell which way it’s oriented?
1. 36y2 − x2 + 72y + 6x+ 31 = 0
(x−32
)2 − (y+11/3
)2= 1
2. y2 − 10y − 8x+ 49 = 0
(y − 5)2 = 4(2)(x− 3)
3. y2 + 2x2 + 4y + 8x+ 11 = 0
(y + 2)2 +(
x+21/√2
)2= 1
Now, complete squares and put into a form you can graph from.Identify (as applicable) the vertices, foci, axes, directrixes,asymptotes.
Warm up
Recall, to graph a conic function, you want it in the formparabola: (x− x0)2 = 4p(y − y0) or (y − y0)2 = 4p(x− x0),
ellipse:(x−x0a
)2+(y−y0
b
)2= 1,
hyperbola:(x−x0a
)2 − (y−y0b
)2= 1 or
(y−y0a
)2 − (x−x0b
)2= 1
Without doing any algebraic manipulation, decide whether each ofthe following conic sections is a parabola, ellipse, or hyperbola.Can you tell which way it’s oriented?
1. 36y2 − x2 + 72y + 6x+ 31 = 0(x−32
)2 − (y+11/3
)2= 1
2. y2 − 10y − 8x+ 49 = 0 (y − 5)2 = 4(2)(x− 3)
3. y2 + 2x2 + 4y + 8x+ 11 = 0 (y + 2)2 +(
x+21/√2
)2= 1
Now, complete squares and put into a form you can graph from.Identify (as applicable) the vertices, foci, axes, directrixes,asymptotes.
36y2 − x2 + 72y + 6x+ 31 = 0: Hyperbola!
0 = 36y2−x2 + 72y + 6x+ 31
= 36(y2 + 2y)− (x2 − 6x) + 31
= 36((y + 1)2 − 1)− ((x− 3)2 − 9) + 31
= 36(y + 1)2 − (x− 3)2 + 4.
So 4 = (x− 3)2 − 36(y + 1)2. So
1 = (x− 3)2/4− 9(y + 1)2 =(x−32
)2 − (y+11/3
)2.
center: (3,−1); a = 2, b = 1/3,c =
√22 + (1/3)2 =
√37/3 ≈ 2.03; ;
vertices: (3± 2,−1);foci: (3± c,−1); (note: very close to the vertices!)
asymptotes: y + 1 = ±(1/32
)(x− 3)
5 10
-2
-1
36y2 − x2 + 72y + 6x+ 31 = 0: Hyperbola!
0 = 36y2−x2 + 72y + 6x+ 31
= 36(y2 + 2y)− (x2 − 6x) + 31
= 36((y + 1)2 − 1)− ((x− 3)2 − 9) + 31
= 36(y + 1)2 − (x− 3)2 + 4.
So 4 = (x− 3)2 − 36(y + 1)2. So
1 = (x− 3)2/4− 9(y + 1)2 =(x−32
)2 − (y+11/3
)2.
center: (3,−1); a = 2, b = 1/3,c =
√22 + (1/3)2 =
√37/3 ≈ 2.03; ;
vertices: (3± 2,−1);foci: (3± c,−1); (note: very close to the vertices!)
asymptotes: y + 1 = ±(1/32
)(x− 3)
5 10
-2
-1
36y2 − x2 + 72y + 6x+ 31 = 0: Hyperbola!
0 = 36y2−x2 + 72y + 6x+ 31
= 36(y2 + 2y)− (x2 − 6x) + 31
= 36((y + 1)2 − 1)− ((x− 3)2 − 9) + 31
= 36(y + 1)2 − (x− 3)2 + 4.
So 4 = (x− 3)2 − 36(y + 1)2. So
1 = (x− 3)2/4− 9(y + 1)2 =(x−32
)2 − (y+11/3
)2.
center: (3,−1); a = 2, b = 1/3,c =
√22 + (1/3)2 =
√37/3 ≈ 2.03; ;
vertices: (3± 2,−1);foci: (3± c,−1); (note: very close to the vertices!)
asymptotes: y + 1 = ±(1/32
)(x− 3)
5 10
-2
-1
36y2 − x2 + 72y + 6x+ 31 = 0: Hyperbola!
0 = 36y2−x2 + 72y + 6x+ 31
= 36(y2 + 2y)− (x2 − 6x) + 31
= 36((y + 1)2 − 1)− ((x− 3)2 − 9) + 31
= 36(y + 1)2 − (x− 3)2 + 4.
So 4 = (x− 3)2 − 36(y + 1)2. So
1 = (x− 3)2/4− 9(y + 1)2 =(x−32
)2 − (y+11/3
)2.
center: (3,−1); a = 2, b = 1/3,c =
√22 + (1/3)2 =
√37/3 ≈ 2.03; ;
vertices: (3± 2,−1);foci: (3± c,−1); (note: very close to the vertices!)
asymptotes: y + 1 = ±(1/32
)(x− 3)
5 10
-2
-1
y2 − 10y−8x+ 49 = 0: Sideways parabola! (opens to the right)
0 = y2−10y−8x+49 = (y−5)2−25−8x+49 = (y−5)2−8x+24,
so(y − 5)2 = 8x− 24 = 8(x− 3) = 4(2)(x− 3).
vertex: (3, 5); p = 2; focus: (3 + p, 5); directrix: x = 3− p
5
5
10
y2 − 10y−8x+ 49 = 0: Sideways parabola! (opens to the right)
0 = y2−10y−8x+49 = (y−5)2−25−8x+49 = (y−5)2−8x+24,
so(y − 5)2 = 8x− 24 = 8(x− 3) = 4(2)(x− 3).
vertex: (3, 5); p = 2; focus: (3 + p, 5); directrix: x = 3− p
5
5
10
y2 − 10y−8x+ 49 = 0: Sideways parabola! (opens to the right)
0 = y2−10y−8x+49 = (y−5)2−25−8x+49 = (y−5)2−8x+24,
so(y − 5)2 = 8x− 24 = 8(x− 3) = 4(2)(x− 3).
vertex: (3, 5); p = 2; focus: (3 + p, 5); directrix: x = 3− p
5
5
10
y2 + 2x2 + 4y + 8x+ 11 = 0: Ellipse!
0 = y2 + 2x2 + 4y + 8x+ 11
= (y2 + 4y) + 2(x2 + 4x) + 11
= ((y + 2)2 − 4) + 2((x+ 2)2 − 4) + 11
= (y + 2)2 + 2(x+ 2)2 − 1.
So
1 = (y + 2)2 + 2(x+ 2)2 = (y + 2)2 +
(x+ 2
1/√2
)2
.
center: (−2,−2);a = 1, b = 1/
√2,
c =√
12 − (1/√2)2 = 1/
√2 ≈ 0.71;
vertices: (−2,−2± a);foci: (−2,−2± c);major axis: x = −2
-3 -2 -1
-3
-2
-1
y2 + 2x2 + 4y + 8x+ 11 = 0: Ellipse!
0 = y2 + 2x2 + 4y + 8x+ 11
= (y2 + 4y) + 2(x2 + 4x) + 11
= ((y + 2)2 − 4) + 2((x+ 2)2 − 4) + 11
= (y + 2)2 + 2(x+ 2)2 − 1.
So
1 = (y + 2)2 + 2(x+ 2)2 = (y + 2)2 +
(x+ 2
1/√2
)2
.
center: (−2,−2);a = 1, b = 1/
√2,
c =√
12 − (1/√2)2 = 1/
√2 ≈ 0.71;
vertices: (−2,−2± a);foci: (−2,−2± c);major axis: x = −2
-3 -2 -1
-3
-2
-1
y2 + 2x2 + 4y + 8x+ 11 = 0: Ellipse!
0 = y2 + 2x2 + 4y + 8x+ 11
= (y2 + 4y) + 2(x2 + 4x) + 11
= ((y + 2)2 − 4) + 2((x+ 2)2 − 4) + 11
= (y + 2)2 + 2(x+ 2)2 − 1.
So
1 = (y + 2)2 + 2(x+ 2)2 = (y + 2)2 +
(x+ 2
1/√2
)2
.
center: (−2,−2);a = 1, b = 1/
√2,
c =√12 − (1/
√2)2 = 1/
√2 ≈ 0.71;
vertices: (−2,−2± a);foci: (−2,−2± c);major axis: x = −2
-3 -2 -1
-3
-2
-1
y2 + 2x2 + 4y + 8x+ 11 = 0: Ellipse!
0 = y2 + 2x2 + 4y + 8x+ 11
= (y2 + 4y) + 2(x2 + 4x) + 11
= ((y + 2)2 − 4) + 2((x+ 2)2 − 4) + 11
= (y + 2)2 + 2(x+ 2)2 − 1.
So
1 = (y + 2)2 + 2(x+ 2)2 = (y + 2)2 +
(x+ 2
1/√2
)2
.
center: (−2,−2);a = 1, b = 1/
√2,
c =√12 − (1/
√2)2 = 1/
√2 ≈ 0.71;
vertices: (−2,−2± a);foci: (−2,−2± c);major axis: x = −2
-3 -2 -1
-3
-2
-1
Conics with cross-terms
Again, all conic sections satisfy an equation of the form
Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0
for some constants A,B,C,D,E, F .
Last time, we talked about how to analyze anything where B = 0.
Just like we can shift up/down or left/right to get from somethinglike
(y − 5)2 = 8(x− 3) to (y)2 = 8x,
we can rotate to get from something where B 6= 0 to somethingwhere B = 0.
Conics with cross-terms
Again, all conic sections satisfy an equation of the form
Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0
for some constants A,B,C,D,E, F .
Last time, we talked about how to analyze anything where B = 0.
Just like we can shift up/down or left/right to get from somethinglike
(y − 5)2 = 8(x− 3) to (y)2 = 8x,
we can rotate to get from something where B 6= 0 to somethingwhere B = 0.
Conics with cross-terms
Again, all conic sections satisfy an equation of the form
Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0
for some constants A,B,C,D,E, F .
Last time, we talked about how to analyze anything where B = 0.
Just like we can shift up/down or left/right to get from somethinglike
(y − 5)2 = 8(x− 3) to (y)2 = 8x,
we can rotate to get from something where B 6= 0 to somethingwhere B = 0.
Take the usual x, y-axis
, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).Now, using the angle addition formulas,
x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).Now, using the angle addition formulas,
x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.
Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).Now, using the angle addition formulas,
x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!
In the x, y-frame,x = r cos(φ) y = r sin(φ).
But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).Now, using the angle addition formulas,
x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).
But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).Now, using the angle addition formulas,
x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).But in the x, y-frame, the angle is φ+ θ, and the radius is thesame!
So in the x, y-frame,x = r cos(φ+ θ) y = r sin(φ+ θ).
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
We want to take a point (x, y) and write it in terms of x and y.Easier to do in polar!In the x, y-frame,
x = r cos(φ) y = r sin(φ).But in the x, y-frame, the angle is φ+ θ, and the radius is thesame! So in the x, y-frame,
x = r cos(φ+ θ) y = r sin(φ+ θ).
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ)
= x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),
y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).Combining these equations, we get
x cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ)
= x sin(θ) + y cos(θ).Combining these equations, we get
x cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0
= x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ))
= y.
Take the usual x, y-axis, and rotate it by θ in [0, π/2)to get a new x, y-axis:
x
y
x
y
θ
(x, y)
r φ
x = r cos(φ),y = r sin(φ)
x = r cos(φ+ θ)y = r sin(φ+ θ)
Now, using the angle addition formulas,x = r cos(φ+ θ) = r cos(φ) cos(θ)− r sin(φ) sin(θ) = x cos(θ)− y sin(θ),y = r sin(φ+ θ) = r cos(φ) sin(θ) + r sin(φ) cos(θ) = x sin(θ) + y cos(θ).
Combining these equations, we getx cos(θ) + y sin(θ) = x(cos2(θ) + sin2(θ)) + y · 0 = x, and−x sin(θ) + y cos(θ) = x · 0 + y(sin2(θ) + cos2(θ)) = y.
Change of coordinates for rotating axes:
x
y
x
y
θ
(x, y) = (x, y)
If I rotate the x, y-axis by θ to get new coordinates (x, y), then theconversion between coordinate systems is given by
x = x cos(θ)− y sin(θ) and y = x sin(θ) + y cos(θ)
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ)
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4)
=√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y)
, and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4)
=√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy
=(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)
= 12(x
2−y2) =(
x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2)
=(
x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x.
Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0
, which is equivalent to x = 0, andy = −x is equivalent to x+ y = 0, is equivalent to y = 0.
Also, in the x, y frame, the foci are (x, y) = (±2, 0). Sox = ±2 cos(π/4) + 0 · sin(π/4) = ±
√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0
, andy = −x is equivalent to x+ y = 0, is equivalent to y = 0.
Also, in the x, y frame, the foci are (x, y) = (±2, 0). Sox = ±2 cos(π/4) + 0 · sin(π/4) = ±
√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0
, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0).
Sox = ±2 cos(π/4) + 0 · sin(π/4) = ±
√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4)
= ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2
, andy = ±(2) sin(π/4) + 0 · cos(π/4) = ±
√2.
Example
Show that the function xy = 1 is a hyperbola rotated by π/4.What are the foci and the asymptotes?
Solution. Convert coordinates, using θ = π/4:
x = x cos(π/4)− y sin(π/4) =√22 (x− y), and
y = x sin(π/4) + y cos(π/4) =√22 (x+ y).
So, subbing into xy = 1, we get
1 = xy =(√
22 (x− y)
)(√22 (x+ y)
)= 1
2(x2−y2) =
(x√2
)2−(
y√2
)2.
In the x, y frame, the asymptotes are y = ±x. Buty = x is equivalent to x− y = 0, which is equivalent to x = 0, and
y = −x is equivalent to x+ y = 0, is equivalent to y = 0.Also, in the x, y frame, the foci are (x, y) = (±2, 0). So
x = ±2 cos(π/4) + 0 · sin(π/4) = ±√2, and
y = ±(2) sin(π/4) + 0 · cos(π/4) = ±√2.
Graphing xy = 1:
In summary, the function xy = 1 is the function(
x√2
)2−(
y√2
)2rotated by θ = π/4. In the x, y-frame, the asymptotes are x = 0and y = 0. The foci are (
√2,√2) and (−
√2,−√2).
^𝑥
^𝑦
𝑥
𝑦
(Compare to y = 1/x!)
Graphing xy = 1:
In summary, the function xy = 1 is the function(
x√2
)2−(
y√2
)2rotated by θ = π/4. In the x, y-frame, the asymptotes are x = 0and y = 0. The foci are (
√2,√2) and (−
√2,−√2).
^𝑥
^𝑦
𝑥
𝑦
(Compare to y = 1/x!)
Graphing xy = 1:
In summary, the function xy = 1 is the function(
x√2
)2−(
y√2
)2rotated by θ = π/4. In the x, y-frame, the asymptotes are x = 0and y = 0. The foci are (
√2,√2) and (−
√2,−√2).
^𝑥
^𝑦
𝑥
𝑦
(Compare to y = 1/x!)
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.Collect “like terms” in x and y.The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.Collect “like terms” in x and y.The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.Collect “like terms” in x and y.The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.
Collect “like terms” in x and y.The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.Collect “like terms” in x and y.
The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
Hypothesis: for a conic sectionAx2 +Bxy + Cy2 +Dx+ Ey + F = 0, there is some rotationalchange of coordinates in which the xy term is missing.
In the previous example, we were explicitly given the right angle torotate by to eliminate the xy-term.
Big question: How could we find this in general?
Strategy:Take a generic θ and plug the change of coordinates
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
in to the generic conic section equation.Collect “like terms” in x and y.The coefficient of xy will be in terms of θ, A, B, etc.. Solve for θso that, in this new equation, the coefficient of xy is 0.
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:
−A cos(θ) sin(θ)−A sin(θ) cos(θ)+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)
−A sin(θ) cos(θ)+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)
−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)
+C sin(θ) cos(θ) + C cos(θ) sin(θ)= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ)
+ C cos(θ) sin(θ)= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))
= B cos(2θ) + (C −A) sin(2θ).So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0
, i.e. tan(2θ) = B/(A− C) .
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
= A(x cos(θ)− y sin(θ))(x cos(θ)− y sin(θ))+B(x cos(θ)− y sin(θ))(x sin(θ) + y cos(θ))
+ C(x sin(θ) + y cos(θ))(x sin(θ) + y cos(θ))
+D(x cos(θ)− y sin(θ)) + E(x sin(θ) + y cos(θ)) + F
Coefficient of xy:−A cos(θ) sin(θ)−A sin(θ) cos(θ)
+B cos(θ) cos(θ)−B sin(θ) sin(θ)+C sin(θ) cos(θ) + C cos(θ) sin(θ)
= B(cos2(θ)− sin2(θ)) + (C −A)(2 cos(θ) sin(θ))= B cos(2θ) + (C −A) sin(2θ).
So we want θ such that
B cos(2θ) + (C −A) sin(2θ) = 0, i.e. tan(2θ) = B/(A− C) .
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?
Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.
Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4
, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1.
So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.
You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
Rotating by θ eliminates the xy terms whentan(2θ) = B/(A− C).
What θ do you want to rotate by to graph each of the followingconic sections?Example: 8x2 − 4xy + 4y2 + 1 = 0.Solution: A = 8, B = −4, C = 4, so tan(2θ) = 4/(8− 4) = 1. So2θ = π/4. So θ = π/8.You try:
1. x2 − xy + y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator.
Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)−y sin(θ), y = x sin(θ)+y cos(θ), tan(2θ) = B/(A−C).
Of course, plugging in things like θ = π/8 into sin(θ) and cos(θ)isn’t fun without a calculator. Draw a triangle for cos(2θ) and usethe half angle formulas!
cos2(θ) = 12(1 + cos(2θ)) sin2(θ) = 1
2(1− cos(2θ))
Example: If tan(2θ) = 1, then 2θ is the angle of the triangle
1
1c
2θ
c =√12 + 12 =
√2
cos(2θ) = 1/√2 =√2/2
cos(θ) =√
12(1 +
√2/2)
sin(θ) =√
12(1−
√2/2)
So
x = x√
12 (1 +
√22 )−y
√12 (1−
√22 ), y = x
√12 (1−
√22 )+y
√12 (1 +
√22 )
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
x = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ)
tan(2θ) = B/(A− C).cos2(θ) = 1
2(1 + cos(2θ)) sin2(θ) = 12(1− cos(2θ))
You try: Find the change of coordinates that will eliminate the xyterm in each of the following conic sections.
1. x2 − xy − y2 − 2 = 02. 4x2 − 4xy + 7y2 − 24 = 03. 2x2 + 4
√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
Identifying the conic section
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
In the case we did last time, where B = 0, we identified the casesshape condition if B = 0:
parabola A = 0 or C = 0
AC = 0
ellipse A and C same signs
AC > 0
hyperbola A and C different signs
AC < 0Just like we did for calculating the coefficient of xy after changingcoordinates, we could calculate all the other coefficients after achange in coordinates
0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F.If we did, we could also show that the discriminants before andafter the change are the same, i.e.
B2 − 4AC = (B)2 − 4AC.But if we chose θ so that B = 0 (like we’ve been doing), then weget
B2 − 4AC = −4AC.
Identifying the conic section
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
In the case we did last time, where B = 0, we identified the casesshape condition if B = 0:
parabola A = 0 or C = 0 AC = 0ellipse A and C same signs AC > 0
hyperbola A and C different signs AC < 0Just like we did for calculating the coefficient of xy after changingcoordinates, we could calculate all the other coefficients after achange in coordinates
0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F.
If we did, we could also show that the discriminants before andafter the change are the same, i.e.
B2 − 4AC = (B)2 − 4AC.But if we chose θ so that B = 0 (like we’ve been doing), then weget
B2 − 4AC = −4AC.
Identifying the conic section
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
In the case we did last time, where B = 0, we identified the casesshape condition if B = 0:
parabola A = 0 or C = 0 AC = 0ellipse A and C same signs AC > 0
hyperbola A and C different signs AC < 0Just like we did for calculating the coefficient of xy after changingcoordinates, we could calculate all the other coefficients after achange in coordinates
0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F.If we did, we could also show that the discriminants before andafter the change are the same, i.e.
B2 − 4AC = (B)2 − 4AC.
But if we chose θ so that B = 0 (like we’ve been doing), then weget
B2 − 4AC = −4AC.
Identifying the conic section
0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
In the case we did last time, where B = 0, we identified the casesshape condition if B = 0:
parabola A = 0 or C = 0 AC = 0ellipse A and C same signs AC > 0
hyperbola A and C different signs AC < 0Just like we did for calculating the coefficient of xy after changingcoordinates, we could calculate all the other coefficients after achange in coordinates
0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F.If we did, we could also show that the discriminants before andafter the change are the same, i.e.
B2 − 4AC = (B)2 − 4AC.But if we chose θ so that B = 0 (like we’ve been doing), then weget
B2 − 4AC = −4AC.
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0
B2 − 4AC = 0
ellipse AC > 0
B2 − 4AC < 0
hyperbola AC < 0
B2 − 4AC > 0
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).
Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0
B2 − 4AC = 0
ellipse AC > 0
B2 − 4AC < 0
hyperbola AC < 0
B2 − 4AC > 0
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0
B2 − 4AC = 0
ellipse AC > 0
B2 − 4AC < 0
hyperbola AC < 0
B2 − 4AC > 0
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0
B2 − 4AC = 0
ellipse AC > 0
B2 − 4AC < 0
hyperbola AC < 0
B2 − 4AC > 0
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0 B2 − 4AC = 0
ellipse AC > 0 B2 − 4AC < 0
hyperbola AC < 0 B2 − 4AC > 0
Putting it together: we start with0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F,
and convert to0 = A(x)2 +Bxy + C(y)2 +Dx+ Ey + F,
usingx = x cos(θ)− y sin(θ), y = x sin(θ) + y cos(θ).
We choose θ such that B = 0, i.e. tan(2θ) = B/(A− C).Then since B2 − 4AC = (B)2 − 4AC, we get
−4AC = B2 − 4AC.
Using our table from before, we can classify the conic sectionaccording to the following table:
shape condition
parabola AC = 0 B2 − 4AC = 0
ellipse AC > 0 B2 − 4AC < 0
hyperbola AC < 0 B2 − 4AC > 0
You try
Identify the kind of conic section given by each of the followingequations:
1. x2 − xy − y2 − 2 = 0
2. 4x2 − 4xy + 7y2 − 24 = 0
3. 2x2 + 4√3xy + 6y2 − 9x+ 9
√3y − 63 = 0
Graphing 0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
1. If B 6= 0, change coordinates using x = x cos(θ)− y sin(θ),y = x sin(θ) + y cos(θ), where tan(2θ) = B/(A− C).Tip: Use a triangle and the half-angle formulas to calculatecos(θ) and sin(θ).
2. With your new equation0 = A(x)2 + C(y)2 +Dx+ Ey + F,
get into one of the canonical forms and calculate all theimportant features of the graph.(If B = 0, then A = A, C = C, etc..)
3. Sketch the graph by first sketching the rotated axes and thengraphing theˆcurve on those axes.
4. If needed, convert features of the graph (like vertices, foci,asymptotes, etc.) back into x, y-coordinates using
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ).
You try: Sketch a graph of 4x2 − 4xy + 7y2 − 24 = 0.
Graphing 0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
1. If B 6= 0, change coordinates using x = x cos(θ)− y sin(θ),y = x sin(θ) + y cos(θ), where tan(2θ) = B/(A− C).Tip: Use a triangle and the half-angle formulas to calculatecos(θ) and sin(θ).
2. With your new equation0 = A(x)2 + C(y)2 +Dx+ Ey + F,
get into one of the canonical forms and calculate all theimportant features of the graph.(If B = 0, then A = A, C = C, etc..)
3. Sketch the graph by first sketching the rotated axes and thengraphing theˆcurve on those axes.
4. If needed, convert features of the graph (like vertices, foci,asymptotes, etc.) back into x, y-coordinates using
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ).
You try: Sketch a graph of 4x2 − 4xy + 7y2 − 24 = 0.
Graphing 0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
1. If B 6= 0, change coordinates using x = x cos(θ)− y sin(θ),y = x sin(θ) + y cos(θ), where tan(2θ) = B/(A− C).Tip: Use a triangle and the half-angle formulas to calculatecos(θ) and sin(θ).
2. With your new equation0 = A(x)2 + C(y)2 +Dx+ Ey + F,
get into one of the canonical forms and calculate all theimportant features of the graph.(If B = 0, then A = A, C = C, etc..)
3. Sketch the graph by first sketching the rotated axes and thengraphing theˆcurve on those axes.
4. If needed, convert features of the graph (like vertices, foci,asymptotes, etc.) back into x, y-coordinates using
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ).
You try: Sketch a graph of 4x2 − 4xy + 7y2 − 24 = 0.
Graphing 0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
1. If B 6= 0, change coordinates using x = x cos(θ)− y sin(θ),y = x sin(θ) + y cos(θ), where tan(2θ) = B/(A− C).Tip: Use a triangle and the half-angle formulas to calculatecos(θ) and sin(θ).
2. With your new equation0 = A(x)2 + C(y)2 +Dx+ Ey + F,
get into one of the canonical forms and calculate all theimportant features of the graph.(If B = 0, then A = A, C = C, etc..)
3. Sketch the graph by first sketching the rotated axes and thengraphing theˆcurve on those axes.
4. If needed, convert features of the graph (like vertices, foci,asymptotes, etc.) back into x, y-coordinates using
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ).
You try: Sketch a graph of 4x2 − 4xy + 7y2 − 24 = 0.
Graphing 0 = Ax2 +Bxy + Cy2 +Dx+ Ey + F
1. If B 6= 0, change coordinates using x = x cos(θ)− y sin(θ),y = x sin(θ) + y cos(θ), where tan(2θ) = B/(A− C).Tip: Use a triangle and the half-angle formulas to calculatecos(θ) and sin(θ).
2. With your new equation0 = A(x)2 + C(y)2 +Dx+ Ey + F,
get into one of the canonical forms and calculate all theimportant features of the graph.(If B = 0, then A = A, C = C, etc..)
3. Sketch the graph by first sketching the rotated axes and thengraphing theˆcurve on those axes.
4. If needed, convert features of the graph (like vertices, foci,asymptotes, etc.) back into x, y-coordinates using
x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ).
You try: Sketch a graph of 4x2 − 4xy + 7y2 − 24 = 0.